Proof of the Cardinality of the power set of $\Bbb Q$, i realize the continuum hypotheses if taken says there no cardinality between $|\Bbb N|$ and $|\Bbb R $ i also know that the cardinality of a power set is strictly larger than than the cardinality of the original set. I am assuming this means that it ends up being that the cardinality of the power set of any countable infinite set is the same as the cardinality of $\Bbb R$. Is there a nice map that shows this?
Proof of the Cardinality of the power set of $\Bbb Q$
elementary-set-theory
Best Answer
I suspect that you meant to ask whether - assuming $\mathsf{CH}$ - $\mathbb{R}$ has the same cardinality as the powerset of any countably infinite set.
The answer is yes, and in fact this has nothing to do with $\mathsf{CH}$. The points are these:
Without $\mathsf{CH}$ we can show $\vert\mathcal{P}(\mathbb{N})\vert=\vert\mathbb{R}\vert$.
Also without $\mathsf{CH}$ we can show that the powerset operation preserves equicardinality. Suppose $\vert A\vert=\vert B\vert$. Let $f:A\rightarrow B$ be a bijection, and consider the map $$g:\mathcal{P}(A)\rightarrow\mathcal{P}(B): X\mapsto \{f(x): x\in X\}.$$ It's a good exercise to show that $g$ is also a bijection. So if two sets have the same cardinality (e.g. if they're both countably infinite) then their powersets also have the same cardinality.
As a particular consequence, if $A$ is countably infinite then $\vert A\vert=\vert\mathbb{N}\vert$ and so $\vert\mathcal{P}(A)\vert=\vert\mathcal{P}(\mathbb{N})\vert=\vert\mathbb{R}\vert$. And again, none of the above involves $\mathsf{CH}$.